Cred ca R=8 cm
Tr EOF avem [tex]EO ^{2} +FO ^{2}=EF ^{2} [/tex]=> TR EOF dr is(R.T.P)=> m(<EOF)=90=>m(arc EF)=90
b)[tex]A _{EOF} = \frac{OE\cdot OF\cdot sin\ \textless \ O}{2} \\ 32 sin O=32 \\ sin \ \textless \ O=1=\ \textgreater \ m(\ \textless \ O)=90=\ \textgreater \ m(arcEF)=90[/tex]
